Q.Use the following data to calculate ΔlatticeH° for NaBr.
ΔsubH° for sodium metal = 108.4 kJ mol^-1
Ionization enthalpy of sodium = 496 kJ mol^-1
Electron gain enthalpy of bromine = -325 kJ mol^-1
Bond dissociation enthalpy of bromine = 192 kJ mol^-1
ΔfH° for NaBr(s) = -360.1 kJ mol^-1

Lattice energy $\Delta H_{lattice}$ is the energy required to completely separate the ions of one mole of an ionic crystal to infinite distance -- for NaCl, $NaCl(s)\rightarrow Na^+(g)+Cl^-(g)$, $\Delta H_{lattice}=+788$ kJ mol$^{-1}$. Because this cannot be measured directly (no calorimeter watches a crystal fly apart into gaseous ions), it is calculated indirectly via the Born-Haber cycle (Max Born and Fritz Haber), which applies Hess's law to the formation of an ionic solid $MX$ by breaking it into five measurable steps: sublimation of the metal ($\Delta H_1$, $M(s)\rightarrow M(g)$), dissociation of the halogen ($\Delta H_2$, $\tfrac{1}{2}X_2(g)\rightarrow X(g)$), ionisation of the metal ($\Delta H_3$, $M(g)\rightarrow M^+(g)$), electron affinity of the halogen ($\Delta H_4$, $X(g)\rightarrow X^-(g)$), and the lattice enthalpy $U$ itself ($M^+(g)+X^-(g)\rightarrow MX(s)$), summed as $\Delta H_f=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+U$. Since every …